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Vilka [71]
3 years ago
8

What is the square root of 9537153​

Mathematics
2 answers:
Anton [14]3 years ago
4 0

Step-by-step explanation:

bruh all u gotta do is googgle that its not that hard. like frl ∞

Dmitriy789 [7]3 years ago
3 0

3088.22813276

hope this helps, you could've entered this into a calculator for next time.

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The length of the longer leg of a right triangle is 6 inches more than twice the length of the shorter leg. The length of the hy
Studentka2010 [4]

9514 1404 393

Answer:

  15 in, 36 in, 39 in

Step-by-step explanation:

The Pythagorean theorem tells us that for short side x, the relation is ...

  (2x +9)² = (2x +6)² +x²

  4x² +36x +81 = 4x² +24x +36 +x²

  x² -12x -45 = 0 . . . . . subtract the left-side expression

  (x -15)(x +3) = 0 . . . . factor

  x = 15 . . . . . . . . . . . the positive value of x that makes a factor zero

The side lengths of the triangle are 15 inches, 36 inches, and 39 inches.

6 0
3 years ago
Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua
goblinko [34]
<h2>Hello!</h2>

The answer is:  There is a total of 5.797 gallons pumped during the given period.

<h2>Why?</h2>

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

D(t)=\frac{5t}{1+3t}

So, the integral will be:

\int\limits^4_0 {\frac{5t}{1+3t}} \ dx

So, integrating we have:

\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx

Performing a change of variable, we have:

1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}

Then, substituting, we have:

\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du

\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )

\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du

\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]

Reverting the change of variable, we have:

\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]

Then, evaluating we have:

\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0
3 years ago
Which of the following best defines 3 to the power of 2 over 3?
Paha777 [63]
3 to the power of 3 is 9 or 3x3=9

9/3=3 


7 0
3 years ago
Read 2 more answers
What are the coordinates of point G?
SSSSS [86.1K]

Answer:

D. (4,9)

Step-by-step explanation:

5 0
3 years ago
The measurement of a side of a square is found to be 12 centimeters. Estimate the maximum allowable percent error in measuring t
Marat540 [252]
 <span>If x= side of the square, and A = area of square, then  

A = x^2 

dA/dx = 2x, or 

dA = 2x*dx, where dx is the differential in x (same as error) 


dA/A = 2x*dx/(x^2) = 2*dx/x = 2*(0.05cm)/(15cm) = 0.00667 = 0.667% 

for b): 

if dA/A = 0.025 = 2*dx/x, then 

dx/x=(0.025)/2 = 0.0125 = 1.25 % 

 I hope I helped. :)</span>
3 0
3 years ago
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